Chapter 3.1, Problem 6E

To determine

To calculate: The derivative of the given function at the given value of a using the definition $f^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Answer to Problem 6E

The required value is 2.

Explanation of Solution

Given information:

The function: $f\left(x\right)=x^2+4$ , $a=1$

Calculation:

The given function is $f\left(x\right)=x^2+4$ .

Write the given definition using the above function as:

  $f^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{\left(x^2+4\right)-\left(a^2+4\right)}{x-a}$

Substitute 1 for a in the above limit and simplify.

  $\begin{array}{c}{f}^{\prime }\left(1\right)=\underset{x\to 1}{\lim }\left[\frac{\left(x^2+4\right)-\left(1^2+4\right)}{x-1}\right]\\ =\underset{x\to 1}{\lim }\left[\frac{x^2+4-5}{x-1}\right]\\ =\underset{x\to 1}{\lim }\left[\frac{x^2-1}{x-1}\right]\\ =\underset{x\to 1}{\lim }\left[\frac{\left(x-1\right)\left(x+1\right)}{x-1}\right]\end{array}$

Simplify the limit further.

  $\begin{array}{c}{f}^{\prime }\left(1\right)=\underset{x\to 1}{\lim }\left[\frac{\left(x-1\right)\left(x+1\right)}{x-1}\right]\\ =\underset{x\to 1}{\lim }\left[\frac{\cancel{\left(x-1\right)}\left(x+1\right)}{\left(\cancel{x-1}\right)}\right]\\ =\underset{x\to 1}{\lim }\left(x+1\right)\end{array}$

Substitute the 1 for x in the limit and simplify.

  $\begin{array}{c}{f}^{\prime }\left(1\right)=1+1\\ =2\end{array}$

Hence, the required value is 2.